1. Sep 19, 2011

Andru10

The uncertainty principle says that you can't simultaneously measure position and momentum with arbitrary precision. So you can measure one at a time t1 and the other at a time t2 with t2 > t1, thus not measuring both simultaneously, but relativity tell us that there exists a frame of reference in which these events are simultaneous and therefore an observer in that frame of reference will notice the measurements being carried out on both with arbitrary precision at the same time. How does one solve this apparent paradox?

2. Sep 19, 2011

DrChinese

For one, the order of measurements will be the same in all reference frames.

3. Sep 19, 2011

Andru10

I think I understand now. Because of the Lorentz transforms, the values measured for position and momentum will, in the frame of reference in which the measurements are simultaneous, be such so that Heisenberg's uncertainty relation is satisfied.

4. Sep 19, 2011

nonequilibrium

Hello, the statement you claim to be the uncertainty principle is an uncertainty principle, but not the one that is usually mathematically proven; this type of UP has so far only been argued on heuristic grounds, as far as I'm aware, and is rather controversial.

But more importantly, your statement about relativity theory is certainly wrong: you cannot make any two events happen at the same time, only if they're separated by a space-like interval, but if two events are causally connected (which measurements should be), then the separation is time-like. Okay judging by your original post the words "space-like" and "time-like" will probably sound unfamiliar at this stage. I suggest to read http://en.wikipedia.org/wiki/Spacetime or to simply believe that the statement "there is always an inertial reference frame where two certain events happen at the same time" is not always right.

5. Sep 19, 2011

Andru10

I am familiar with time-like and space-like intervals, but I considered the 2 measurements to not be causally linked and therefore space-like. Also, that version of the UP has been proven... the operators for position and momentum do not commute.

6. Sep 19, 2011

nonequilibrium

Correct me if I'm wrong, but the proof you're talking about, leads to the $\sigma_x \sigma_p \geq \textrm{constant}$ UP where the sigmas are the standard deviations of the respective probability densities, so this UP is talking about what manifests itself if you do N identically-prepared measurements: the plot of the measurements gives you the original probability distributions and all the UP is talking about, is the standard deviations of those densities.

But anyway, on the relativity issue: what would two causally unconnected measurements look like?

And as a third issue: shouldn't you use QFT when taking relativity into account?

(And my apologies for assuming you didn't know time-like and space-like intervals, didn't mean anything by it.)

7. Sep 19, 2011

BruceW

The strict definition of a 'measurement' in QM will depend on who you ask, due to large controversy. This makes it a difficult concept to come to terms with.

You can be sure that it is not possible to have an eigenstate of both momentum and position.

8. Sep 19, 2011

Andru10

Well all UPs lead to a relationship like that. You are correct, but I'm trying to relate to the principles of QM, principles which are assumed to be correct (in the Copenhagen interpretation). One principle states that observables are operators in the complex Hilbert space and another states that the measured values for the observables are eigenvalues of these operators. It can be proven that in order to simultaneously measure 2 observables the respective operators need to commute and it is proven that for position and momentum (along the same axis) the operators do not commute.

I don't know what two causally unconnected measurements look like, perhaps this is my error in judgement. I simply ment that the 2 measurements need not be causally linked (it doesn't matter in which order you perform them or where).

I'm not sure how to use QFT in this situation.

No apologies necessary :).

9. Sep 19, 2011

edpell

If the distance between them is greater than the distance light can travel in t2-t1 then it is space-like. But of course this is impossible since the particle you are measuring must travel faster than the speed of light. It is time-like regardless of wither there is cause and effect in play.

10. Sep 19, 2011

BruceW

If you make the assumption that non-unitary collapse doesn't happen (ie ManyWorldsInterpretation), then QM is a local, real phenomena. So therefore it is impossible to make two causally unconnected measurements. (Or even to affect the wavefunction in any way which is faster than light).

I'm not sure how to resolve the apparent paradox using Copenhagen Interpretation, but it probably hinges on exactly how the definition of 'simultaneous measurements' is made.

11. Sep 19, 2011

mathfeel

In any given frame, suppose you measure the particle to be in position x1 at time t1. You would have localized the wave function. Now if you measure again at t2, it cannot have spread out (I am thinking Gaussian wave packets) from x1 much more than c (t2 - t1), so it has very small probability over |x_2 - x_1| > c (t2 - t1).

So the two measuring events are in fact time-like connected. There is no other frame in which they are simultaneous (or t2' < t1' for that matter). I could be remembering this incorrectly: the Green's function for a free relativistic particle G(t2-t1, x2-x1), which gives you exactly what you are asking for: the probability amplitude that if you measure at (t1, x1) then again at (t2, x2), is exponentially small for space-like time interval.

Exponentially small is non-zero. So one'd still argue violation of causality. Indeed, this is only resolved when you include an anti-particle and its propagator cancels the amplitude for the amplitude for the particle outside the lightcone. There is a discussion of this in Sec.-2.4 of Peskin and Schroeder's QFT book.

Last edited: Sep 19, 2011
12. Sep 19, 2011

BruceW

There are instances where you can make measurements on particles far away (in Copenhagen Interpretation).
For example, have a spin-up, spin-down pair of particles, and separate them by a very large distance, then measure the spin of one, which will cause the collapse of the other particle into the other spin state.

Edit: Although, I don't know how you could use a similar set-up to make a measurement on momentum, rather than angular momentum of a far-away particle.

13. Sep 19, 2011

PAllen

If you are measuring position and momentum of the 'same thing' at two different times, the measurements are necessarily timelike. The measurements occur at two times on the world line of the thing measured. This order will never change, not matter what the motion of the observer is. If, instead, they occur for the same time on the "thing's" world line, they are simultaneous for the purposes of the uncertainty principle.

In short: Dr. Chinese answer was a complete answer to the question.

14. Sep 19, 2011

Andru10

I am satisfied that the measurements are time-like. Thank you all!

15. Sep 20, 2011

Demystifier

If the separation between the two measurements is spacelike, then for one of these two measurements there will no be particle there to measure it. That's because the particle cannot be faster than light.